Link budget — path loss, EIRP, G - T, Eb - N0: Question bank

3.6.25 · D5 · Physics › Spacecraft Structures & Systems Engineering › Link budget — path loss, EIRP, G - T, Eb - N0

Yeh ek conceptual question bank hai link budget topic ke liye. Yahaan koi bhaari arithmetic nahi hai — woh kaam worked-example pages ka hai. Balki, har item ek misconception ya boundary case ko probe karta hai jo maths quietly invite karta hai. Reveal tab hi padho jab tum apna jawab zor se bol chuke ho.

Neeche har reveal mein tumhe reasoning milti hai, kabhi ek bare "haan" ya "nahi" nahi. Agar tum kyun reconstruct kar sako — toh concept tumhara hai.

True or false — justify

Har statement ya to true hai ya false. Reveal batata hai kaunsa, aur physical reason.

Transmitter power

P_{t}

ko double karne se EIRP watts mein double ho jaata hai.

True. EIRP

= P_{t} \cdot G_{t}

linear units mein, toh

P_{t}

ko 2 se scale karne par EIRP bhi 2 se scale hota hai — lekin dB mein yeh sirf

+ 3

dB ka change hai, kyunki

10 lo g_{10} (2) \approx 3

Distance

r

ko double karne se path loss decibels mein quadruple ho jaata hai.

False.

r

ko double karne se linear power ratio chaar guna ho jaata hai (inverse-square), lekin dB mein yeh sirf

20 lo g_{10} (2) \approx 6

dB add hota hai, multiply nahi — dB values add hote hain, kabhi multiply nahi karte.

Path loss ka daromedar is baat par hai ki tum dono ends par kaun se antennas choose karte ho.

False. Free-space path loss

L_{p} = (4 π r / λ)^{2}

mein sirf distance aur wavelength hain; antenna gains budget mein alag terms hain. Dekho Free-Space Path Loss Derivation.

Ek higher frequency link hamesha same distance ke liye zyada free-space path loss suffer karta hai.

True.

L_{p} \propto (r / λ)^{2}

aur

λ = c / f

, toh bada

f

matlab chhota

λ

matlab bada

L_{p}

— higher frequency, ek fixed dish size ke liye zyada spreading loss.

System noise temperature

T_{sys}

badhane se G/T figure of merit barhta hai.

False.

G / T = G_{r} - 10 lo g_{10} (T_{sys})

; zyada

T_{sys}

zyada subtract karta hai, G/T ko gira deta hai. Kam noise temperature behtar hoti hai. Dekho Noise Temperature and Noise Figure.

Ek fixed received carrier power

C

ke liye, data faster bhejne se

E_{b} / N_{0}

kam ho jaata hai.

True.

E_{b} / N_{0} = (C / N_{0}) - 10 lo g_{10} (R_{b})

, toh bit rate

R_{b}

badhane par zyada subtract hota hai, har bit ko milne wali energy shrink ho jaati hai.

EIRP sirf transmitter ki property hai, signal ke antenna se nikalne se pehle measure ki jaati hai.

True. EIRP sirf transmit power aur transmit antenna gain combine karta hai; iska channel ya receiver se koi lena-dena nahi.

C / N_{0}

aur

E_{b} / N_{0}

ek hi quantity hai jo alag units mein express ki gayi hai.

False.

C / N_{0}

(dB-Hz) ek carrier-to-noise-density ratio hai jo data rate se independent hai;

E_{b} / N_{0}

dimensionless hai aur

R_{b}

par depend karta hai. Dono mein

10 lo g_{10} (R_{b})

ka fark hota hai.

Spot the error

Har line mein ek flawed statement ya reasoning hai. Reveal galti ka naam batata hai.

"Received power EIRP minus path loss hai, toh bade receive antennas koi help nahi kar sakte."

Receive antenna gain

G_{r}

ek alag additive term hai;

C = EIRP - L_{p} + G_{r}

. Bada dish

G_{r}

badhata hai aur received power seedha boost karta hai. Dekho Antenna Gain and Effective Aperture.

"Kyunki path loss 279 dB hai aur EIRP sirf 34 dBW hai, toh link impossible hai."

Tum EIRP ko akele

L_{p}

se compare nahi kar sakte — tumhe receive

G / T

term bhi add karni hai aur Boltzmann's constant (

- 228.6

dBW/K/Hz) bhi subtract karna hai. Yeh bade positive contributions link ko bacha lete hain.

"G/T mein gain dBi mein hai aur temperature Kelvin mein, toh main 25 K system ke liye seedha 25 subtract kar deta hoon."

Tum

10 lo g_{10} (T_{sys}) = 10 lo g_{10} (25) \approx 14

dB subtract karte ho, raw number 25 nahi. Temperature logarithmically enter hoti hai.

"Bit error rate aadhi karni hai toh bas transmit power double karo."

BER

E_{b} / N_{0}

ke saath girta hai lekin steeply aur non-linearly (threshold ke paas roughly exponentially), toh

+ 3

dB power bump BER ko orders of magnitude gira sakta hai — relationship proportional nahi hai. Dekho Modulation Schemes (BPSK, QPSK).

"BPSK ki jagah QPSK use karne se same

E_{b} / N_{0}

par data rate free mein double ho jaati hai."

QPSK 2 bits per symbol pack karta hai, lekin same bit error rate ke liye BPSK aur QPSK ko same

E_{b} / N_{0}

chahiye — faayda spectral efficiency ka hai, free performance nahi. Extra bits energy terms mein "free" nahi hain.

"Boltzmann's constant ek loss hai, toh hum budget mein

+ 228.6

dB subtract karte hain."

Hum

228.6

add karte hain (yaani negative

- 228.6

dBW/K/Hz ko subtract karte hain). Kyunki

k

bahut chhota hai,

10 lo g_{10} (k)

ek bada negative number hai, aur ise remove karna help karta hai

C / N_{0}

ko.

Why questions

Reason explain karo, sirf fact nahi.

Hum poora link budget linear watts ki jagah decibels mein kyun kaam karte hain?

Kyunki gains aur losses linear units mein multiply hote hain lekin dB mein add hote hain; ek dozen terms ka addition bahut asaan aur kam error-prone hai, compared to 30 orders of magnitude tak phele bade aur chhote numbers ko multiply karne ke.

Path loss

1/ r^{2}

ke saath kyun badhta hai jab ki koi wave absorb nahi kar raha?

Wahi total power

4 π r^{2}

area ke ek ever-larger spherical shell par phail jaata hai; energy kho nahi jaati, bas dilute ho jaati hai — power density purely geometry se

1/ r^{2}

ke roop mein girta hai.

Digital links ke liye plain signal-to-noise ratio ki jagah

E_{b} / N_{0}

preferred metric kyun hai?

E_{b} / N_{0}

data rate ko normalise kar deta hai, toh yeh seedha bit error rate predict karta hai aur Shannon-Hartley Theorem information-theory limit se cleanly connect hota hai, alag speeds par chal rahe links ko fairly compare karne deta hai.

Forward error correction ek link ko kam

E_{b} / N_{0}

par kyun close hone deta hai compared to uncoded link ke?

FEC redundant bits add karta hai jo decoder ko errors correct karne deta hai, toh yeh noisier channel tolerate kar sakta hai — raw bit rate ko "coding gain" ke liye trade karta hai required

E_{b} / N_{0}

mein kaafi dB ka. Dekho Forward Error Correction.

Deep-space ground stations cryogenic amplifiers se bahut kam noise temperatures kyun pursue karte hain?

Kyunki

N_{0} = k T_{sys}

noise floor set karta hai, aur ek door probe ke fixed weak signal ke liye,

T_{sys}

aadha karna seedha

G / T

aur

C / N_{0}

badhata hai — yeh ek aakhiri lever hai jab signal khud fixed ho. Dekho Deep Space Network (DSN).

Edge cases

Boundary aur degenerate scenarios jo formulas quietly assume karte hain.

Path-loss formula

r \to 0

par kya predict karta hai, aur yeh unphysical kyun hai?

L_{p} = (4 π r / λ)^{2} \to 0

, matlab zero loss (ya "gain") zero range par — lekin formula assume karta hai far-field spherical wave, toh yeh antenna ke close near field mein simply apply nahi hota.

Agar receive antenna gain

G_{r}

0 dBi (isotropic receiver) ho, toh kya link phir bhi kaam karta hai?

Kar sakta hai — 0 dBi matlab gain factor 1 (koi help nahi, koi nuksan nahi); term budget mein sirf

+ 0

contribute karta hai. Link tabhi survive karta hai jab EIRP, kam

T_{sys}

G / T

, aur kam data rate path loss compensate kar sakein.

E_{b} / N_{0}

ka kya hota hai jab data rate

R_{b} \to 0

E_{b} / N_{0} = (C / N_{0}) - 10 lo g_{10} (R_{b})

unbounded badhta jaata hai jab

R_{b} \to 0

, kyunki har vanishingly rare bit saari carrier energy hoard kar leta hai — yeh reflect karta hai kyun bahut slow links enormous distances par close ho sakte hain.

Shannon limit par, minimum

E_{b} / N_{0}

kya hai jo ek link theoretically tolerate kar sakta hai?

Approximately

- 1.6

dB (

ln 2 \approx 0.693

linear terms mein); is se neeche koi bhi coding scheme, chahe kitni bhi clever ho, arbitrarily low error rates achieve nahi kar sakti. Dekho Shannon-Hartley Theorem.

Agar do spacecraft equidistant hain lekin ek half wavelength use karta hai, toh kiska path loss zyada hai, aur dB mein kitna?

Half-wavelength (double-frequency) link ka loss zyada hai,

20 lo g_{10} (2) \approx 6

dB se, kyunki

L_{p} \propto (1/ λ)^{2}

aur

λ

aadha karne se linear loss chaar guna ho jaata hai.

Recall Self-test: teen "quiet subtractions"

Un teen terms ke naam batao jo link budget mein subtract hoti hain (ya subtraction jaisi lagti hain) aur ek trap har ek ke liye. Answer ::: (1) Path loss $L_{p}$ — trap: yeh bahut bada hai lekin antenna gains aur $G / T$ ise offset karte hain. (2) $10 lo g_{10} (R_{b})$ in $E_{b} / N_{0}$ — trap: faster data $E_{b} / N_{0}$ gira deta hai. (3) $10 lo g_{10} (k)$ Boltzmann — trap: yeh ek bada negative number hai jise tum effectively add back karte ho.